Molar mass Na2CO3= 105.99g/mol Mass of Na2CO3= 0.2123g Moles of Na2CO3= 0.2123g/105.99g/mol Moles of Na2CO3= 2.003 x 10-3 moles Mole-to-Mole Ratio 1 Na2CO3: 2 HCl Moles of HCl= 2 x 2.003 x 10-3moles Moles of HCl= 4.006 x 10-3 Molarity of HCl= (4.006 x 10-3)/0.04304 Molarity of HCl= 0.09308M Moles of HCl= 0.09308M x 0.03073L Moles of HCl= 2.8601 x 10-3 moles Moles of Na2CO3 = (2.860 x 10-3 moles) / 2 Moles of Na2CO3 = 1.4301 x 10-3 Mass of Na2CO3== 1.4301 x10-3
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Standardize NaOH Average NaOH after 3 trials was 0.6578 M NaOH Moles of HCL Molarity of HCL (average between bottle 1 & 1A) Grams of HCL Grams of Water 4.319g HCL = 4.7181 g H20 Moles of Water in HCL Solution Moles of Water added Moles of ester initially present moles of ester Moles of alcohol initially present Moles of acid present Moles of carboxylic acid present‚ moles of alcohol formed‚ moles of ester disappeared‚ moles of
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sulfur dioxide gas in flask B. moles ethane 10/30 0.33 mol moles SO2 0.33 mol mass SO2 0.33 64 21 g 2 What mass of nitric oxide‚ NO‚ is present in a 2.5 L flask at a pressure of 100 kPa and 0ºC? mole NO 2.5/22.71 0.1101 mol mass NO 0.1101 30 3.3 g 3 Carbon monoxide burns in oxygen according to the equation: 2CO(g) O2(g) → 2CO2(g) Calculate the volume of oxygen required at 25°C and 100 kPa for the combustion of 28 g of carbon monoxide. moles CO 28/28 1 mol
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Equation HX + NaOH NaX + H2O 1 mole 1 mole 1 mole 1 mole Aim To analyze the Vitamin C from a rival company and compare with that of my company and find out the best value for the money spent by the consumer. Controlled Variable * Mass of the tablet and hereby vitamin C in rival company tablets * Consistency of components in the Vitamin C tablet * Concentration of NaOH Apparatus 1. Burette (50ml) 2. Conical flask (50ml) 3. Vitamin C tablet (1 tablet
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084×10-1molL-1 =1.492 x 10-3 moles Number of moles of A.S.A can be found by the difference between the mean number of moles of NaOH added for hydrolysis and the mean number of moles of HCl. Avg.vol=sample volumes# of samples Avg. vol = (42.83 + 43.40) 2 = 43.12 mL # of moles of NaOH = Mean vol of NaOH x Molar Conc. Of NaOH = 4.312 x 10-2L ×0.09085molL-1 =3.917 x 10-3 moles # of mol A.S.A = mean # of moles NaOH - mean # moles HCl =
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the number of moles. 3. Find the simplest ratio of moles. Question 1. Mass of elements in g. 2. Number of moles of each element. 3. Simple ratio of elements. Practice problems * A 2.765 g sample of lead oxide was heated in a stream of hydrogen gas and completely converted to elemental lead with a mass of 2.401g. What is the empirical formula of the oxide? Mass of oxygen = 2.765 – 2.401 = 0.364 g Number of moles of oxygen = 0.364/16 = 0.02275 Number of moles of lead = 2.401/20719
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Titration and Volumetric Analysis The purpose of this experiment is to determine the [NaOH] of a solution by titrating it with standard HCl solution‚ to neutralize a known mass of an unknown acid using the NaOH solution as a standard‚ to determine the moles of NaOH required to neutralize the unknown acid‚ and to calculate the molecular mass of the unknown acid. Procedure: Part A: Standarized 0.10M HCl solution and unknown NaOH solution were poured into two beakers. The burets were then filled with
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Chemical Reaction – Lab Christian Lecce Mr. Ribarich Wednesday‚ February 20st‚ 2013 Purpose To determine the mass of copper formed when excess aluminum is reacted with a given mass of a copper salt (Copper Chloride dihydrate)‚ and the mole-to-mole ratio between the reactant and the product of a chemical reaction. Apparatus * 150ml beaker * Stirring rod * Ruler * Hotplate * Tweezers * 50ml graduated cylinder Materials * Copper (II) chloride dehydrate
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Processing of Uncertainties | No of moles of Mg in reaction= 0.12524.31= 0.00514 mol (to 3sf)No of moles of Oxygen in reaction= 0.04216.00= 0.00263 mol (to 3sf)No of moles of Mg : No of moles of O= 2:1 ∴ Empirical formula of the compound = Mg2O | Percentage Uncertainty of No of moles of Mg in reaction= 0.0020.125 ×100%= 1.60% (to 3sf)Absolute uncertainty of No of moles of Mg in reaction= 0.00514 × 1.60100= 0.00008 mol (to 5dp)Percentage Uncertainty of No of moles of O in reaction= 100% × 0.0050.042=
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The initial moles of copper used is .1574 moles. I calculated the mass of the oxygen by subtracting the mass of the crucible before the reaction and the afterwards. Then‚ in order to determine the moles of oxygen used during the reaction‚ I divided the mass of oxygen by its molar mass of 16.00 g/mol. The moles of oxygen was determined to be .1574. Afterwards‚ by dividing moles of copper over moles of oxygen‚ I determined that the molar relationship between
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